Mathematics, 18.11.2020 22:40 keyshawn437

Please Help ASAP AP calculus Due in 1 day

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Answer from: pizzaboy62

(D) $\displaystyle \frac{12}{1 + ln8}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Algebra I

FunctionsFunction Notation

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integration

Integration Rule [Fundamental Theorem of Calculus 2]: $\displaystyle \frac{d}{dx}[\int\limits^x_a {f(t)} \, dt] = f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt$

Step 2: Differentiate

Chain Rule: $\displaystyle f'(x) = \frac{d}{dx} \bigg[ \int\limits^{x^3}_1 {\frac{1}{1 + ln(t)}} \, dt \bigg] \cdot \frac{d}{dx}[x^3]$ Integration Rule - Fundamental Theorem of Calculus 2: $\displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot \frac{d}{dx}[x^3]$ Basic Power Rule: $\displaystyle f'(x) = \frac{1}{1 + ln(x^3)} \cdot 3x^{3 - 1}$ Simplify: $\displaystyle f'(x) = \frac{3x^2}{1 + ln(x^3)}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle f'(2) = \frac{3(2)^2}{1 + ln(2^3)}$ Exponents: $\displaystyle f'(2) = \frac{3(4)}{1 + ln(8)}$ Multiply: $\displaystyle f'(2) = \frac{12}{1 + ln(8)}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

Same length is comparing because it’s saying its the same and not different

Answer from: Quest

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

= $\sqrt{(9-6)^{2} +(12-6)^{2} } \\=\sqrt(45)=3\sqrt(5)$

side j'k'=distance between j' and k'

= $\sqrt{(3-2)^{2} +(2-4)^{2} } \\=sqrt(5)$

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl

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